$\int (x+6)(3x-4)\,dx=$ $+C$
The integrand is the product of two functions: $x+6$ and $3x-4$. Although it is tempting to take the product of their integrals, this would not work. $\int f(x)\cdot g(x)\,dx\neq\int f(x)\,dx \cdot \int g(x)\,dx$ Instead, what we should do is expand the parentheses so we get a nice polynomial. $\int (x+6)(3x-4)\,dx=\int (3x^2+14x-24)\,dx$ Now we can integrate using the reverse power rule, the sum rule, and the constant multiple rule for indefinite integrals. $\begin{aligned} &\phantom{=}\int (x+6)(3x-4)\,dx \\\\ &=\int (3x^2+14x-24)\,dx \\\\ &= 3\int x^{2}\,dx +14\int x\,dx -24\int 1\,dx \\\\ &=3\dfrac{x^3}{3} +14\dfrac{x^2}{2} -24\dfrac{x^1}{1}+C \\\\ &=x^3 +7 x^2 -24 x +C \end{aligned}$ In conclusion, $\int (x+6)(3x-4)\,dx=x^3 +7 x^2 -24 x +C$